#### Question

A goods train leaves a station at 6 p.m., followed by an express train which leaved at 8 p.m. and travels 20 km/hour faster than the goods train. The express train arrives at a station, 1040 km away, 36 minutes before the goods train. Assuming that the speeds of both the train remain constant between the two stations; calculate their speeds.

#### Solution

Let the speed of goods train be x km/hr. So, the speed of express train will be (x + 20) km/hr.

Distance = 1040 km

We know

Time taken by good train to cover a distance of 1040 km = `1040/x` hrs

Time taken by express train to cover a distance of 1040 km = `1040/(x + 20)` hrs

It is given that the express train arrives at a station 36 minutes before the goods train. Also the express train leaves the station 2 hours after the goods train. This means that the express train arrives at the station (36/60 + 2) hrs = 13/5 hrs before the good train.

Therefore we have

`1040/x - 1040/(x + 20) = 13/5`

`(1040x + 20800 - 1040x)/(x(x + 20)) = 13/5`

`20800/(x^2 + 20x) = 13/5`

`1600/(x^2 + 20x) = 1/5`

`x^2 + 20x - 8000 = 0`

`(x - 80)(x + 100) = 0`

x = 80, -100

Since the speed cannot be negative So x = 80

Thus the speed of goods train is 80 km/hr and the speed of express train is 100 km/hr